Here is the construction of $\Bbb CP^n$ explained in Hatcher book (p. 7):
It is also possible to obtain $\Bbb CP^n$ as a quotient space of the disk $D^{2n}$ under the identifications $v\sim \lambda v$ for $v\in \partial D^{2n}$, in the following way. The vectors in $\Bbb S^{2n+1}\subset \Bbb C^{n+1}$ with last coordinate real and nonnegative are precisely the vectors of the form $(w, \sqrt{1-|w|^2})\in \Bbb C^n\times \Bbb C$ with $|w|\leq 1$. Such vectors form the graph of the function $w\mapsto \sqrt{1-|w|^2}$. This is a disk $D_+^{2n}$ bounded by the sphere $\Bbb S^{2n-1}\subset\Bbb S^{2n+1}$ consisting of vectors $(w, 0)\in \Bbb C^n\times \Bbb C$ with $|w|= 1$. Each vector in $\Bbb S^{2n-1}$ is equivalent under the identifications $v\sim \lambda v$ to a vector in $D_+^{2n}$, and the latter vector is unique if its last coordinate is nonzero. If the last coordinate is zero, we have just the identifications $v\sim \lambda v$ for $v\in \Bbb S^{2n-1}$.
I don't understand why the dimension of the above disk is $2n$? (and of course the dimension of the $\Bbb S^{2n-1}$) I think it (dimension of above disk) should be $2n+1$ because it is the upper hemisphere of $\Bbb S^{2n+1}$. What is the wrong?
It is in fact a little confusing. The point is to understand $S^{2n+1}$. Obviously we have $$S^{2n+1} = \{ (z_1,\ldots,z_{n+1}) \in \mathbb C^{n+1} \mid \sum_{i=1}^{n+1} \lvert z_i \rvert^2 = 1 \} .$$ Hatcher considers the subset $$S = \{(z_1,\ldots,z_{n+1}) \in S^{2n+1} \mid \text{Im}(z_{n+1}) = 0, \text{Re}(z_{n+1}) \ge 0\} .$$ But $S' = \{(z_1,\ldots,z_{n+1}) \in S^{2n+1} \mid \text{Im}(z_{n+1}) = 0 \}$ is nothing else than a copy of $S^{2n} \subset \mathbb R^{2n+1} = \mathbb C^n \times \mathbb R$.
Thus $S$ is not the upper hemisphere of $S^{2n+1} \subset \mathbb R^{2n+2}$, but a copy of the upper hemisphere of $S^{2n} \subset \mathbb R^{2n+1}$. And this is clearly homeomorphic to a $2n$-dimensional disk.
More formally, the disk $D^{2n} \subset \mathbb C^n$ is homeomorphic to $S$ because it is the graph of the function $\phi : D^{2n} \to \mathbb C, \phi(w) = \sqrt{1-\lvert w \rvert}$.